Unformatted text preview: g ( x ) on all of R , we use that it is periodic. Thus, for any x,y ∈ R so that | x-y | < δ (assuming without loss that δ < 1) we can substract 2 πn for some n ∈ Z so that x-2 πn and y-2 πn are in [-2 π, 2 π ]. Then, | sin x-sin y | = | sin( x-2 πn )-sin( y-2 πn ) | < ε since | ( x-2 πn )-( y-2 πn ) | < δ . To show nonuniform continuity of fg on R , we shall use that sin x/x → 1 as x → and Theorem 5.4.2. Indeed, let x n = 2 πn and u n = 2 πn + 1 n . Then x n-u n → 0. x n sin x n = 0 → 0. But by the sine addition formula, u n sin u n = (2 πn + 1 /n ) sin 1 /n . But 2 πn sin 1 /n → 2 π . Thus for n suﬃciently large, | fg ( x n )-fg ( u n ) | = (2 πn + 1 /n ) sin 1 /n ≥ 1 / 2 . Hence by the Nonuniform Continuity Criterion, fg is not uniformly continous. 1...
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- Spring '10
- Calculus, Sin, Continuous function, Metric space, Uniform continuity